Cone of Fire

Lots of questions emerge from this Fast Draw game, and one that I've been pondering is what the cone of fire has to be to hit a 2 foot target at 21 feet.  I'm a practical sort, so I asked the question to a geometry teacher of my acquaintance, via Facebook.

Hey, Mandy! I have a geometry question: I have an isosceles triangle, the base is 2 feet (24 inches), the long legs are 21 feet. What is the angle of the acute angle (pointy end)?

In her inimitable style, she answered me within the hour.

the vertex angle is 5.4588052735573928698235743338306 degrees; the base angles are 84.541194726442607130176425666169 degrees

 I'd say that's fairly precise (carried to 30 decimal places, but it answers my question.  About 5.4 degrees.  That means that if we aim dead-square at the bullseye, our aim can be off by about 2.7 degrees in any direction and we'll still hit the 2-foot target at 21 feet.  Any more than that is a clean miss.

The question might best be illustrated by this graphic.  The lens shortened the distance, but you can get an idea of the concept.

That's not terribly hard when you're using the sights, but when you're shooting from the hip that's a mighty small angle to get a hit. To put that another way, if you draw your revolver and point it at the target, your muzzle can only vary 1/10th of an inch from dead on, or you'll miss the target.  That's fairly precise.  (According to my math, X = 0.0952 inches).

Thanks, ;Mandy.